MULTIPLE CHOICE FINAL REVIEW
Units 1-9
1. The image below shows red blood cells in solution. Label the type of solution found in each
sample, from left to right.
a. Isotonic, hypertonic, hypotonic
b. Hypotonic, isotonic, hypertonic
c. Hypertonic, hypotonic, isotonic
d. Hypertonic, isotonic, hypotonic
Questions #2-3 refer to the following situation:
In the 1890s, hunting reduced the northern elephant seal population to only 20 individuals. Due to
conservation efforts, their population size today is above 30,000.
2. How would the genetic diversity of the northern elephant seal population compare to the
genetic diversity of the southern elephant seal population, which was not heavily hunted?
a. The northern elephant seal would have much less genetic variation than the southern
elephant seal.
b. The southern elephant sea would have much less genetic variation than the northern
elephant seal.
c. The two populations would have similar genetic variation.
d. The southern elephant seal would have stable genetic variation because it is in HardyWeinberg equilibrium.
3. This is an example of what biological principle?
a. Founder effect
b. Allopatric speciation
c. Bottleneck effect
d. Microevolution
4. While eukaryotic cells are generally larger than prokaryotic cells, metabolic requirements place
upper limits on how large eukaryotic cells can become. Which of the following statements does
not accurately describe this?
a. If a cell is too large, there will not be enough surface area for the exchange or
elimination of oxygen, nutrients, and wastes.
b. The surface area to volume ratio must remain as large as possible.
c. As multicellular organisms grow in size, they do not have larger cells; rather, they have a
larger number of cells.
d. As a cell increases in size, surface area grows faster than volume.
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5. Starch, glycogen, and cellulose are all composed of long chains of glucose molecules. While
starch and glycogen are often branched, cellulose is always straight due to hydrogen bonding
between neighboring strands. How is this structural difference advantageous for the function of
cellulose?
a. Cellulose is held tightly together in order to offer support in cell walls.
b. Cellulose is used as long-term energy storage, held tightly together to take up minimal
space.
c. Cellulose is a structural polysaccharide, used as support in exoskeletons.
d. The lack of branching allows cellulose to be easily hydrolyzed, enabling it to be used as
quick energy.
6. Animal muscle is composed of a variety of different types of proteins. Muscles use ATP as the
source of energy to create work. When an animal is stressed, ATP is depleted quickly in the
muscles. This results in an environment that has a low pH and a high temperature. Under these
conditions, muscle proteins may denature. If denaturation occurs, which of the following is not
interrupted in the protein molecules?
a. α helices
b. disulfide bridges
c. peptide bonds
d. hydrophobic interactions
7. Using the graph comparing photosynthetic rate and absorption of different wavelengths of light,
what can be concluded?
a. Carotenoids absorb the narrowest range
wavelengths of light
b. Chlorophyll a absorbs wavelengths at both
ends of the spectrum
c. The photosynthetic rate at 550 nm is the
highest
d. Chlorophyll b is associated with the greatest
rate of photosynthesis.
Questions #8-9 refer to the following information:
Prairie dogs are a type of rodent that inhabit the grasslands of
central and western North America. These burrowing animals
live in large colonies and are an important component of the
ecosystem. The prairie dog foraging technique increases plant
diversity. Their burrows enable water to find underground
aquifers, thereby conserving water. The act of burrowing into
the ground mixes the soil, creating richer growing conditions. The open spaces maintained by prairie
dogs increase the animal diversity found in the ecosystem.
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8. Prairie dogs are a keystone species for the Great Plains ecosystem. Which of the following is also
true of this animal?
a. They have a low biomass in the ecosystem.
b. They are at the top of the food chain.
c. Removal of the prairie dog will not have a large impact on the ecosystem.
d. They do not experience competition.
9. To determine the effect of prairie dog on the Great Plains ecosystem, you create an exclusion
experiment. You maintain several plots of land that have distinct borders to them. In one of the
plots, you exclude prairie dogs. Which of the following is not a plausible result in the exclusion
plot?
a. Plant diversity decreases.
b. Prolonged drought.
c. Herbivore populations increase.
d. Nutrient limitation in the soil.
10. All living organisms use DNA and RNA as their genetic code. Humans and roundworms, in fact,
have 25% of their DNA in common while humans and chimpanzees have 98% in common. What
type of evidence of evolution is this?
a. Comparative embryology
b. Fossil record
c. Molecular homology
d. Biogeography
11. Organisms have evolved a variety of adaptations that allow them to survive their environments.
One aspect of survivorship involves the trade-off between the quantity and quality of offspring
produced. What is a true statement regarding this type of adaptation?
a. Organisms who follow r-selection favor quality over quantity.
b. Organisms who follow K-selection experience high fecundity.
c. Producing numerous offspring consumes energy but ensures that at least some will
grow into reproducing adults.
d. Organisms who favor quality have low parental involvement.
12. An RNA strand has the following sequence: 5’ – UAGCUGAC – 3’. How many molecules of water
did it require to create this strand of nucleotides?
a. 7
b. 8
c. 9
d. 10
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Questions #13-14 refer to the following image of a fruit fly oocyte.
Bicoid protein concentration
Highest
Lowest
A
D
B
C
13. Based on the diagram, where would the anterior part of the fruit fly develop?
a. A
b. B
c. C
d. D
14. Bicoid protein is a type of egg-polarity gene that determines the body axes of the fruit fly. What
is another name for this type of gene?
a. Maternal effect gene
b. Oncogene
c. Molecular coordinate gene
d. Homeotic gene
15. Cyclin and kinases are molecules that aid in cell cycle regulation. Which of the following is true
concerning these molecules?
a. They function independently of each other.
b. Cyclin regulates the G1 and G2 checkpoints while kinases regulate the metaphase
checkpoint.
c. Kinases are only active when paired with cyclin.
d. The concentration of cyclin within cells remains constant.
16. Which of the following cannot be determined from looking at a karyotype?
a. The sex of the individual.
b. Down syndrome, in which there is an extra copy of chromosome 21.
c. Klinefelter syndrome, in which there is an abnormal number of sex chromosomes.
d. The loci of specific genes.
17. Snakes evolved from ancestors who had the ability to walk. Some snakes, in fact, still have small
pelvis and leg bones that are remnants from this ancestor. What are these called?
a. Vestigial structure
b. Homologous structure
c. Sympatric structure
d. Molecular homology
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Questions #18-19 refer to the following cellular components:
I.
II.
III.
IV.
V.
Plasma membrane
Nucleus
Peroxisomes
Ribosomes
Chloroplasts
18. Ameobas and Chlamydomonas are two types of protists. Ameobas engulf food while
Chlamydomonas are phototrophs, using the sun’s energy to create food. Which cellular
component is not found in both types of protists?
a. I
b. III
c. IV
d. V
19. Which cellular components are found in all organisms?
a. I and II
b. II and IV
c. I and IV
d. I, II, and IV
20. Chemosynthetic autotrophs and photosynthetic autotrophs are similar in that they both capture
free energy to produce organic compounds. Which of following correctly contrasts
photosynthesis and chemosynthesis?
a. In photosynthesis oxygen gas is the final electron acceptor while in chemosynthesis
hydrogen sulfide is the final electron acceptor.
b. In chemosynthesis water is a final product while in photosynthesis carbon dioxide is a
final product.
c. Photosynthetic organisms require solar radiation while chemosynthetic organisms
require electromagnetic radiation.
d. Chemosynthetic organisms capture energy from inorganic molecules while
photosynthetic organisms capture energy present in sunlight.
21. Although they are both nucleic acids, RNA and DNA carry out different functions. Which
statement describes a similarity between the two macromolecules?
a. They are both composed of 5-carbon sugars.
b. They are double stranded.
c. They share two purines and two pyrimidines.
d. They are self-replicating.
22. When an enzyme is noncompetitively inhibited, which of the following occurs?
a. The enzyme is able to increase its activity.
b. The active site will change shape.
c. The active site will be occupied by the inhibitor molecule.
d. Increasing substrate concentration will increase inhibition.
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23. Mendel’s law of independent assortment states that alleles for different traits are inherited
independently from each other. To what type of gene does this law not apply?
a. Codominant genes
b. Dihybrid genes
c. Linked genes
d. Incompletely dominant genes
Questions #24-26 refer to the following situation.
An experiment was conducted to determine the factors affecting aerobic respiration in a bacterium.
Under aerobic conditions, the rate of carbon dioxide production was measured at 5 minute intervals. In
each treatment, a different molecule was added to the bacterial solution. A graph of the results is shown
below.
1.2
1
0.8
Control
0.6
Citrate Added
0.4
ATP Removed
0.2
AMP Added
0
0
5
10
15
20
24. Which of the following best explains the above results?
a. ATP and citrate are allosteric inhibitors in the cellular respiration process while the
presence of AMP stimulates respiration.
b. ATP and AMP stimulate respiration, while citrate inhibits this process.
c. Citrate stimulates the production of CO2 by simulating fermentation in the bacterial
cells.
d. The presence of ATP and AMP have little effect on the process of cellular respiration.
25. When oxygen is removed from the environment, CO2 continues to be produced at low levels.
What type of bacterium is being used in the experiment?
a. Obligate anaerobe
b. Obligate aerobe
c. Facultative anaerobe
d. Oxidative anaerobe
26. How many treatments are being used in this experiment?
a. 1
b. 3
c. 4
d. 8
27. Which of the following is not evidence for the existence of LUCA?
a. Living organisms use a universal genetic code.
b. Proteins are built using the same 20 amino acids.
c. Nucleic acids are used as the energy source.
d. Lipid bilayers are used in cellular membranes.
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28. In an experiment, scientists transfer fibroblasts, a type of connective tissue cell, to two cultures
with different growth mediums. Both growth mediums contain glucose, amino acids, salts, and
antibiotics. Culture #2, however, also includes protein called PDGF. After incubating the cultures
at 37⁰ C for 24 hours, culture #1 shows no cell division while the cells of culture #2 have
extensively divided. What can the scientists conclude about PDGF?
a. It is capable of binary fission.
b. It is a growth factor for fibroblasts.
c. It is only effective at high temperatures.
d. It prevents anchorage dependence.
29. The difference between a negative feedback loop and a positive feedback loop is that in a
negative feedback loop the new signal will _______________ the stimulus.
a. Slow down
b. Accelerate
c. Promote
d. Reverse
30. Willows are a type of deciduous tree found in temperate climates. Although plants generally
reproduce sexually, if a stem from a willow is placed in moist soil, a new tree will grow. This new
tree will be genetically identical to the parent tree. This is an example of what type of
reproduction?
a. Binary fission
b. Budding
c. Fragmentation
d. Spores
Question #31 refers to the image below depicting the genes found on chromosome 4 of fruit flies.
Bent wings
Eyeless
Shaven bristles
Sparkling eyes
31. Which of the following gene pairs would have the highest recombination frequency?
a. Bent wings and sparkling eyes
b. Shaven bristles and sparkling eyes
c. Bent wings and eyeless
d. Eyeless and shaven bristles
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32. Which of the following is an example of a positive feedback loop?
a. High body temperature is sensed by neurons, blood vessels in the skin dilate, and sweat
glands are activated, lowering the body temperature.
b. Low blood glucose levels triggers the hydrolysis of glycogen, releasing glucose into the
blood, raising the blood glucose level.
c. Bleeding triggers the release of enzymes that activate platelets in the blood. Activated
platelets release chemicals to activate more platelets, leading to a blood clot.
d. Partially digested food in the intestines stimulates the liver to reduce the amount of
digestive enzymes.
33. A pentose is the general name for a 5-carbon sugar. The carbonyl groups of some pentoses are
located at the end of the carbon chain. These molecules are called aldopentoses. The carbonyl
groups of other pentoses are located within the carbon chain. These molecules are called
ketopentoses. Regardless of carbonyl group location, all pentoses will have the following
molecular formula:
O
-C – H
Carbonyl Group
a.
b.
c.
d.
C10H20O10
C5H3O4
C10H8O9
C5H10O5
34. Before RNA polymerase can begin transcription, transcription factors must first bind to the
promoter region. Of what is this an example?
a. Allosteric regulation
b. Alternative splicing
c. The specificity of enzymes
d. Regulation of gene expression
35. Having a widow’s peak (W) is dominant over not having one (w). John has a widow’s peak, but
his mother does not. What is a correct statement about the widow’s peak gene of John’s father?
a. He must be homozygous recessive for the treat.
b. He must have at least one dominant allele.
c. He must be heterozygous for the trait.
d. He is incompletely dominant for the trait.
36. Steroids, a type of lipid, include the sex hormones that are synthesized and stored in the testes
and ovaries. The testes and ovaries, therefore, are rich in what organelle?
a. Ribosomes
b. Rough ER
c. Mitochondria
d. Smooth ER
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Question #37 refers to the following information:
An experiment was conducted to measure the reaction rate of the human salivary enzyme alphaamylase. Ten mL of a concentrated starch solution and 1.0 mL of alpha-amylase solution were
placed in a test tube. The test tube was inverted several times to mix the solution and then
incubated at 25 degrees C. The amount of product (maltose) present was measured every 10
minutes for an hour. The results are given in the table below.
37. Why was a change in the reaction rate observed after 30 minutes?
a. The rate increases as the substrate concentration declines.
b. The rate decreases as the substrate concentration declines.
c. The product is increased.
d. The enzyme is buffered.
38. Many herbicides act on different stages of photosynthesis. If a particular herbicide disrupts the
action of ATP synthase, which is also true?
a. More ATP than usual is formed.
b. Chlorophyll cannot absorb photons.
c. The citric acid cycle cannot be completed.
d. Cyclic electron flow can still occur.
39. Which one is not a type of physical pathway between adjacent cells to aid in communication and
transfer of substances?
a. Tight junctions
b. Quorum junctions
c. Plasmodesmata
d. Adhering junctions
40. Innate immunity is general immunity that all animals have. Which of the following is not part of
innate immunity?
a. Skin
b. Mucus
c. Antibodies
d. Inflammation
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Questions #41-43 refer to the following codon table:
Source: http://commons.wikimedia.org/wiki/File:06_chart_pu3.gif
41. Leucine can be coded for by CUU, CUC, CUA, or CUG; however, these same codons will never
code for any amino acid other than leucine. What does this indicate about the genetic code?
a. It is redundant but not ambiguous.
b. It can be recombined to form a variety of products.
c. It is ambiguous but not redundant.
d. It is strictly regulated.
42. This table would be used to decipher the bases on what molecule?
a. DNA template
b. tRNA anticodon
c. mRNA
d. rRNA large subunit
43. Errors during gene expression often result in mutations in proteins. Which of the following
codons would have the largest impact on the resulting protein?
a. UGU
b. UGA
c. UGG
d. AUG
44. Which of the following provides a similarity between stem cells and cancer cells?
a. They are able to reproduce indefinitely.
b. They cause diseases.
c. They are only found in the bone marrow.
d. They replicate via binary fission.
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Questions #45-46 refer to the following scenario:
In pea plants, there are two genes that code for flower color, P and C. Each gene codes for a different
step of flower pigment production. If either gene is homozygous recessive, the flower will be white.
Otherwise, the flower is purple.
45. If a PPCc plant is crossed with a Ppcc plant, what phenotypic ratio will result?
a. 3 purple: 1 white
b. 3 white: 1 purple
c. All white
d. 1 purple: 1 white
46. This situation, in which more than one gene affects one trait, is referred to by what name?
a. Codominance
b. Epistasis
c. Pleiotropy
d. Linked genes
47. Margarine is produced from plant oil that has undergone hydrogenation. In the process of
hydrogenation, most of the carbon double bonds are removed. As a result of hydrogenation, the
melting point of the product is elevated. An example of hydrogenation is illustrated below:
Which of the following statements is incorrect based on this information?
a. Hydrogenation produces trans fats.
b. Hydrogenation creates fats that are healthier than the starting product.
c. Margarine is solid at room temperature.
d. Consumption of hydrogenated fats affects HDL levels.
48. Morphine is a commonly used drug prescribed for severe pain. Patients who use morphine on a
long-term basis often experience a drug tolerance to it. This means that the effect of morphine
slowly lessons and a higher dosage is required to induce the same level of pain relief. One
theory that explains drug tolerance involves varying pathways that lesson the amount of
receptors that will accept the drug. This is an example of what type of process?
a. Antibiotic resistance
b. Upregulation
c. Downregulation
d. Neural incompatibility
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Questions #49-51 refer to the following scenario:
You are given a u-tube filled with two solutions separated at the bottom by a semi-permeable
membrane. You do not know the compositions of the solutions and are asked to determine which
solution has a higher initial solute potential. You decide to observe the u-tube over several hours.
Illustrated below are the initial and final diagrams of the u-tube.
Solution A
Solution B
Source: http://course1.winona.edu/sberg/ILLUST/fig8-10.GIF
49. What is an accurate description of the initial conditions in the u-tube?
a. Solution A is hypertonic compared to Solution B.
b. Solution A is hypotonic compared to Solution B.
c. Solution A and Solution B have reached osmotic equilibrium.
d. Not enough information.
50. What is an accurate description of the final conditions in the u-tube?
a. Solution A is hypertonic compared to Solution B.
b. Solution A is hypotonic compared to Solution B.
c. Solution A and Solution B have reached osmotic equilibrium.
d. Not enough information.
51. It turns out that Solution A had an initial solute potential of -0.25 bar. Which of the following is a
possible initial solute potential for Solution B?
a. 0.25 bar
b. -0.12 bar
c. -0.25 bar
d. -0.58 bar
52. Which of the following is not something that happens during a cell-mediated response?
a. The infected cell is flagged to alert immune cells of the infection.
b. Histamines increase blood flow.
c. Helper T cells attach to anitgens.
d. Cytotoxic T cells destroy infected cell.
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53. James Watson and Francis Crick worked together to determine the structure of DNA. They
created their model of DNA structure based on the x-ray diffraction picture taken by Rosalind
Franklin. Which of the following supports their idea that a purine always pairs with a pyrimidine
in the DNA molecule?
a. A purine plus a pyrimidine is the only pairing whose width matched the measurements
of the x-ray picture.
b. Covalent bonding is only possible between a purine and a pyrimidine.
c. A pyrimidine bonded to a pyrimidine would have created a pairing that was too wide for
their DNA data.
d. In a given DNA molecule, the amount of cytosine molecules present always equaled the
amount of adenine molecules.
Questions #54-55 refer to the following phylogenetic tree:
Source: http://www.vanderbilt.edu
54. The elephant and the crocodile evolved from a common ancestor. Which characteristic do they
have in common?
a. Amniotic egg
b. Subdermal fat stores
c. Gizzard
d. Ability to see UV light
55. When constructing this phylogenetic tree, several different options were possible. The tree with
the least number of evolutionary changes is the one that was chosen as most probable. What is
this called?
a. Least parsimony
b. Modern synthesis
c. Temporal equilibrium
d. Maximum parsimony
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56. Yellow seeds (Y) are dominant to green seeds (y) in pea plants. A yellow seed plant is crossed
with a green seed plant. The F1 generation is composed entirely of yellow seeds. What is the
name of this procedure?
a. Dihybrid cross
b. Testcross
c. Trihybrid cross
d. Chi-squared test
57. The cell cycle is tightly regulated by various mechanisms, such as checkpoints at specific times of
both interphase and the mitotic phase. Which of the following is not an example of a factor
involved in cell cycle control?
a. Density
b. Growth factors
c. Substratum
d. Independent assortment
Questions #58-59 refer to the following karyotype:
Source: http://www.biology.iupui.edu/biocourses/N100/2k2humancsomaldisorders.html
58. The karyotype portrays Edward’s syndrome. Infants born with this condition usually only survive
for a few months. This condition is an example of what type of event?
a. Polyploidy
b. Fragmentation
c. Nondisjunction
d. Monosomy
59. How does this type of event occur?
a. Chromosome duplication errors during interphase
b. Inaccurate separation of chromosomes during meiosis
c. Lack of independent assortment during meiosis
d. Incomplete separation of sister chromatids during mitosis
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Questions #60 refer to the following diagram of the trp operon:
Source: http://commons.wikimedia.org/wiki/File:Lac_Operon.svg
60. In order for tryptophan production to stop, what needs to occur?
a. An inducer needs to block the promoter region, which will inhibit further tryptophan
production.
b. Alternative splicing needs to change the introns present, thereby inhibiting tryptophan
production.
c. Tryptophan needs to bind to the repressor molecule, which in turn will bind to the
operator region, inhibiting further tryptophan production.
d. Tryptophan needs to bind to the promoter region, thereby prohibiting the binding of
RNA polymerase and the production of further tryptophan.
61. Although animals and plants are both able to defend themselves against pathogens, the
methods of defense differ greatly. Which of the following describes a difference between animal
and plant immunity?
a. Each plant cell must defend itself against pathogens.
b. An outer membrane provides an initial line of defense in plant cells.
c. In animals, a signaling cascade results in an immune response.
d. As a last line of defense, infected cells are killed in animals.
62. Scientists who understand the structure of DNA were able to create the gel electrophoresis
procedure. During this procedure, an electrical current forces DNA to move from the negative
end to the positive end. What structural component of DNA allows this to occur?
a. The specific order of bases on the inside of the molecule
b. The negatively-charged backbone of DNA
c. The solubility of deoxyribose
d. The positively-charged phosphate groups
63. Which of these statements is correct regarding the complimentary sequence to this DNA strand:
5’ – GCGA – 3’
a. It will have one uracil.
b. It will start with a 5’ end.
c. It will have one purine.
d. It will have three 2-ring bases.
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Questions #64-65 refer to the following information:
Deer mice are usually found in the woods and are dark brown in color. The deer mice that live in the
Sand Hills of Nebraska, however, have evolved a lighter coloring, the result of a change in one gene.
64. The initial population of deer mice in the Sand Hills was probably dark brown, like all other deer
mice. What was the driving force behind the evolution of this population?
a. Mutation
b. Speciation
c. Equilibrium
d. Predation
65. A scientist decides to move a population of brown deer mice to a habitat composed of a mixture
of sandy areas and dark rocks. After monitoring the population for several generations, she finds
that it is composed of a mixture of very light and very dark colored mice. What type of selection
has occurred?
a. Adaptive
b. Disruptive
c. Directional
d. Stabilizing
66. Towards the end of pregnancy, the uterus becomes more sensitive to the oxytocin, a hormone
that causes uterine contractions. This sensitivity is accomplished by acquiring additional
oxytocin receptors. What is this process called?
a. Quorum sensing
b. Downregulation
c. Allosteric regulation
d. Upregulation
67. Stomata are pores on the surface of plant leaves that allow for gaseous exchange. The chart
below shows the density of stomata on the leaf surfaces of three species of plants. Based on this
information, which statement is incorrect?
Plant
Elodea
Water Lily
Black Walnut
Stomata Density (# of stomata/mm2)
In Upper Epidermis
In Lower Epidermis
0
0
420
0
0
465
a. In Elodea’s aquatic environment, stomata are not used to control gas exchange.
b. Since water lilies live on the surface of water, transpiration occurs from the upper
epidermis because that is where it is in contact with air.
c. Since black walnut trees thrive on land, stomata are located on lower surface to
decrease the amount of water lost in the heat of the sun.
d. Since Elodea does not have any stomata, it does not carry out photosynthesis.
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Questions #68-69 refer to the following diagram:
Source: http://commons.wikimedia.org/wiki/File:Replication_complex_a.png
68. The diagram shows single-stranded binding proteins, helicase, and topoisomerase located near
the replication fork. What common function do these molecules share?
a. The addition of new nucleotides.
b. Unwinding DNA and maintaining it in an unwound position.
c. Proofreading the new nucleotides.
d. Protecting DNA from degradation.
69. Where would one find Okazaki fragments?
a. Behind topoisomerase.
b. In the leading strand.
c. Hydrogen bonded to the single-stranded binding proteins.
d. In the lagging strand.
70. Hemophilia is a recessive sex-linked disorder carried on the X chromosome. The disorder affects
the ability of the blood to clot. A female, who does not have hemophilia, has children with a
male who has hemophilia. They have four children: one male with hemophilia, one healthy
male, one female with hemophilia, and one healthy female. What is true about the mother?
a. She has two normal alleles for the hemophilia gene.
b. She has one normal gene on the X chromosome and one recessive gene on the Y
chromosome.
c. She passed on at least one hemophilia allele to each child.
d. She is a carrier for hemophilia.
71. Bdelloid rotifers are a group of organisms that inhabit a variety of environments. They solely
reproduce via asexual means. Which of the following is not a benefit derived from this lifestyle?
a. They are able to reproduce faster than sexually reproducing organisms.
b. They do not need to spend time and energy attracting mates.
c. The offspring are genetically varied due to random fertilization.
d. They do not use energy to produce sex cells.
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72. Some strains of Staphylococcus aureus are resistant to antibiotics. This bacteria is able to create
an enzyme that prevents the antibiotic from binding to the active site, rendering it useless.
Which of the following is responsible for conferring resistance in bacteria?
a. F plasmids
b. Pili
c. R plasmids
d. Capsule
73. A recent study exposed some seeds to increased amounts of gibberellin, with a control group of
untreated normal seeds. What would be a reasonable result of such a study?
a. Treated seeds grew to show increased phototropism.
b. Treated plants germinated faster.
c. Treated plants grew slower.
d. Treated and untreated plants show no differences in growth patterns.
Question #74 refers to the diagram below:
74. In what direction will the net flow of water occur?
a. Water will move to side A because side A contains the higher concentration of NaCl and
sucrose.
b. Water will move to side B because side B contains the least concentration of NaCl and
sucrose.
c. Water will accumulate on side B when NaCl and sucrose diffuse through the membrane.
d. Water will move toward side A because side A contains the least concentration of NaCl
and sucrose.
75. Which is the correct order of a signal transduction pathway?
a. Reception, transduction, response
b. Reception, response, transduction
c. Transduction, reception, response
d. Transduction, response, reception
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Questions #76-77 refer to the following information:
Planktonic copepods are small crustaceans that float with sea currents in the ocean. Usually only 1-2
mm in length, they feed on phytoplankton also present in the surface waters. Copepods are very
sensitive to contamination by metals and are often used as bioindicators for water pollution.
76. Which of the following is not an abiotic factor that affects planktonic copepods?
a. Copper concentrations
b. Phytoplankton concentrations
c. Water temperature
d. Ocean currents
77. Copepods follow r strategies for reproduction. Which of the following is a true statement
concerning organisms such as these?
a. They value quality over quantity of offspring.
b. They mature at a later age.
c. They live in an unstable environment.
d. Population size is constant.
78. Cladograms are diagrams that show relationships between different organisms. Each branch of a
cladogram indicates a common ancestor. By studying the cladogram below, determine which
two organisms share the greatest genetic relationship.
a.
b.
c.
d.
Flies and moths
Butterflies and bees
Beetles and wasps
Ants and flies
79. The mechanisms of transcription and translation are virtually identical in species from all three
domains, Archaea, Bacteria, and Eukarya. Which of the following hypotheses could be best
supported by this evidence?
a. The mechanisms of transcription and translation presently found in living organisms are
the only mechanisms that could effectively convert heredity information into protein
structures.
b. The mechanisms of transcription and translation are universal processes and therefore
suggest a common ancestor for all forms of life.
c. The similarity in these processes in all organisms suggests that convergent evolution has
occurred.
d. This evidence does not support a hypothesis because the products of transcription and
translation vary widely.
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PSI AP Biology
Final Review
Questions #80-81 refer to the following information:
The hemoglobin protein consists of four subunits. Each subunit contains separate binding sites
for carbon dioxide and oxygen. When one oxygen (or carbon dioxide) binds to a subunit, a
conformation change occurs in the other three subunits allowing them to more readily accept
oxygen (or carbon dioxide). When oxygen is bound to hemoglobin, the protein cannot bind
carbon dioxide. The oxygen-binding site, however, has a higher affinity for carbon monoxide
than it does for oxygen. If carbon monoxide is present, hemoglobin will bind to it instead of
oxygen. This can lead to carbon monoxide poisoning in which the victim suffocates..
80. Although hemoglobin is not an enzyme, it shares several properties with enzymes. Which of the
following is not a similar feature of both hemoglobin and enzymes?
a. Allosteric regulation
b. Competitive inhibition
c. Noncompetitive inhibition
d. Three levels of structure
81. What animal would have a hemoglobin gene most similar to that found in humans?
a. Chimpanzees
b. Hatchet fish
c. Deer mouse
d. Eagle
82. The functions of respiratory and circulatory systems are interrelated. Which statement is not
true?
a. Respiratory systems get rid of the lactic acid that builds up during strenuous exercise.
b. Molecules that enter the body via respiration enter the blood stream of the circulatory
system via diffusion.
c. The respiratory system gets rid of carbon dioxide created during cellular respiration.
d. The circulatory system carries oxygen from the respiratory system to mitochondria for
cellular respiration.
83. Homologous chromosomes are matching chromosomes found in diploid organisms. They share
similar shape and genetic loci. In a diploid organism, where would homologous chromosomes
not be found?
a. Somatic cells
b. Prophase I
c. Interphase prior to mitosis
d. Metaphase II
84. In 1953, Stanley Miller created a simulation of early Earth’s atmosphere. He was able to create
simple compounds, amino acids and hydrocarbons. Which of the following was not one of the
molecules that Miller used in his “atmosphere”?
a. CH4
b. H2S
c. H2O
d. NH3
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Final Review
85. What statement accurately describes the illustration below?
Source: http://commons.wikimedia.org/wiki/File:Phage_Reproduction_Cycle.jpg
a.
b.
c.
d.
A virus is reproducing inside a host cell, ending in the death of the host cell.
A retro virus is invading a bacterial cell.
A virus is coexisting inside a host cell.
A virus is being attacked by restriction enzymes.
86. When making wine, yeast is added to grape juice. As time passes, the yeast consumes all of the
oxygen in the flask, but continues to thrive and produce alcohol. Which of the following
statements best explains this process?
a. Yeast has evolved the ability to uses other electron acceptors in the absence of oxygen.
b. Yeast is a facultative anaerobe, able to switch to fermentation in the absence of oxygen.
c. In the absence of oxygen, yeast participates in a symbiotic relationship with anaerobic
bacteria.
d. Fructose stimulates alcohol production in yeast.
87. In gel electrophoresis, the smallest DNA fragments will travel the farthest. Why does this
consistently occur?
a. Small fragments have less charge on them and therefore travel farther.
b. Small fragments are the first to leave the well and have more time to travel than the
larger fragments.
c. The higher molecular weight of larger fragments makes them sink.
d. Small fragments move more freely through the agar gel.
88. In pea plants, stem length can be either tall or dwarf, with tall (T) being dominant to dwarf (t).
What would be the phenotypic ratio of a monohybrid cross for this trait?
a. 3 tall : 1 dwarf
b. All tall
c. 1 TT : 2 Tt : 1 tt
d. 3 dwarf : 1 tall
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89. All of the following scenarios are examples of barriers that distinguish separate species. Which is
not a postzygotic barrier?
a. Two species of ladybugs live in the same habitat. Although they mate at the same time
of year, their genitals are physically incompatible.
b. Two species of fish inhabit the same stream. They are capable of mating and producing
healthy offspring. The offspring, however, are infertile.
c. Two species of opossums inhabit the same forest. Although they are capable of mating,
no offspring ever survive development.
d. Two species of mice coexist in a forest habitat. They are capable of mating and
producing offspring. The offspring, however, are frail and die young.
Questions #90-91 refer to the following graphs:
90. What can be concluded about P. aurelia and P. caudatum?
a. They both follow exponential growth models.
b. When grown together, P. aurelia competitively excludes P. caudatum.
c. P. aurelia is parasitic to P. caudatum.
d. When grown together, P. caudatum reaches its carrying capacity sooner than P. aurelia.
91. If nutrients and water are continually added to a sample containing both P. caudatum and P.
aurelia, both species survive. When combined with the information in the graphs above,what
does this indicate?
a. P. aurelia competes better for food than P. caudatum.
b. Predation causes P. caudatum populations to crash in the absence of nutrition.
c. In the presence of nutrition, P. caudatum and P. aurelia become facultative symbionts.
d. P. aurelia is a dominant species.
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Final Review
Questions 92-93 refer to the following:
Many scientists postulate that the following four steps brought about the evolution of simple cells:
1) Small, organic molecules were created from inorganic sources.
2) The small molecules combined to form macromolecules.
3) The macromolecules were packaged into protobionts that were able to maintain an
internal chemistry separate from their surroundings.
4) Self-replicating molecules were created and made inheritance possible.
92. The protobionts described in step 3 had a membrane that separated the internal and external
environments. What current macromolecule did this membrane probably resemble?
a. Triglycerides
b. Nucleotides
c. Lipids
d. Proteins
93. What self-replicating molecule do many scientists postulate came first?
a. Proteins
b. Amino acids
c. Glycerol
d. RNA
94. Remoras are marine fish that have a modified fin in the shape of a sucking disc that can attach
to the skins of other marine animals, such as sharks and manta rays. Remoras eat the leftover
food from their hosts’ meals. The host is neither harmed nor benefitted by the remora’s
presence. How would this relationship be characterized?
a. Parasitism
b. Obligate
c. Commensalism
d. Neutralism
95. Which of the following statements does not describe the role of auxin in phototropism?
a. Auxin causes cell elongation.
b. In response to light, auxin moves away from the source.
c. As the concentration of auxin on the darker side of the plant increases, the elongation
of those cells also increases and the plant bends toward the light.
d. In the presence of light, auxin is activated to stunt cell growth.
96. Cyanide is a highly toxic molecule. It bonds to a cytochrome in the electron transport chain
located in the mitochondria, rendering the electron transport chain useless. The ingestion of
cyanide, therefore, results in death due to which of the following?
a. Inability to take in oxygen through the lungs.
b. Build up of lactic acid.
c. Starvation, due to an inability to create glucose.
d. Cellular death from lack of energy.
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Final Review
Answer #97-98 based on an AaBb x aabb cross that results in 20 offspring.
97. In a laboratory setting, this cross is conducted with the following results: 6 AaBb, 5 Aabb, 6 aaBb
and 3 aabb. What is the chi-squared value?
a. 0
b. 0.05
c. 1.67
d. Not enough information.
98. Use the chi-squared chart below and a p value of 0.05 to make a conclusion about the
experiment.
a. The difference between the observed and expected results is statistically significant.
b. You should accept the null hypothesis.
c. The difference between the observed and expected results was not due to random
chance.
d. The experiment should be rerun, using a larger sample size.
99. Spirochaetes are free-living, anaerobic bacteria that contain spiral-shaped cells. Nematodes are
roundworms that are incredibly diverse and, as a group, have adapted to survive in all
environments. Which of the following statements accurately describes the DNA of both groups?
a. The spirochaete genome is smaller than the nematode genome.
b. Nematode DNA is circular.
c. Spirochaete chromosomes are composed of just DNA while nematode chromosomes
also include specialized proteins.
d. Spirochaete DNA is linear.
100. A virus is considered an obligate intracellular parasite. What does this mean?
a. It can only reproduce inside a host cell.
b. It can switch between the lytic and lysogenic cycles.
c. It can use either DNA or RNA as its genetic material.
d. It is dependent upon reverse transcriptase inside the bacterial host cell.
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Final Review
Questions #101-104 refer to the following information:
Nitrogen is the most limiting nutrient for most plants, including Azolla, an aquatic fern. Azolla has
formed a relationship with the cyanobacterium Anabaena azollae. This cyanobacteria fixes atmospheric
nitrogen into a form that can be used by Azolla. In return, Azolla supplies nutrients to the cyanobacteria.
101. How would the relationship between Azolla and Anabaena azollae be classified?
a. Commensalism
b. Competition
c. Parasitism
d. Symbiosis
102. Azolla is used in rice paddies as a natural fertilizer. When paddies are flooded, Azolla are
allowed to grow on the surface. As rice grows, it eventually overshadows Azolla, causing it to
die. How would this be beneficial to the rice?
a. When Azolla dies, the cyanobacteria can transfer usable nitrogen to the rice.
b. The decomposition of Azolla supplies nutrients to the growing rice.
c. The death of Azolla give the rice more room to grow.
d. The oxygen expelled by the Azolla enables the rice to grow.
103. Which of the following is a density independent factor that would affect Azolla?
a. A long winter freeze
b. Predatory fish
c. A similar aquatic plant that inhabits the same space
d. A disease that affects aquatic ferns
104. Large Azolla blooms are prevalent in areas with high phosphorous concentrations. What does
this indicate about Azolla?
a. Phosphorous is toxic to Azolla.
b. Phosphorous is toxic to the cyanobacteria.
c. The cyanobacteria are also able to fix phosphorous.
d. Phosphorous is a limiting nutrient for Azolla.
105. Why is a closed circulatory system more efficient?
a. Muscular movement assists the transport of materials.
b. The transport of nutrients and removal of wastes is accomplished rapidly.
c. Oxygen can enter directly into the blood via diffusion across the skin.
d. Body organs and tissues are continuously in contact with blood.
106. Alzheimer’s disease is a neurodegenerative disease for which there currently is no cure.
Alzheimer’s results from an accumulation of misfolded proteins. In some instances, β pleated
sheets form instead of α helices. At what level of structure is this error occurring?
a. Primary
b. Secondary
c. Tertiary
d. Quaternary
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Final Review
107. Eukaryotes are a diverse group that have evolved several differences from their prokaryotic
ancestors. Which of the following statements does not accurately describe a difference
between eukaryotes and prokaryotes?
a. Eukaryotes have organelles suspended in their cytoplasm.
b. Eukaryotes have smaller cells which enable them to be more efficient.
c. Eukaryotes contain their DNA in a nucleus.
d. Most eukaryotes are multicellular.
108. Due to advances in biotechnology, scientists are able to produce insulin in a laboratory setting.
This enables people who suffer from diabetes to replace the insulin that their bodies are not
producing. During the laboratory procedure, the insulin gene is isolated and then amplified.
What process is used to amplify the insulin gene?
a. Recombinant DNA
b. Splicing
c. Electrophoresis
d. Polymerase chain reaction
Question #109 refers to the image below of cells extracted from the bronchioles of a patient.
Source: http://commons.wikimedia.org/wiki/File:Tripolar_Mitosis_-_bronchial_wash.jpg
109. The cell in the middle of the picture shows tripolar mitosis, indicating that the cell is cancerous.
Tripolar mitosis explains what aspect of cancer?
a. Cancer cells do not respond to cell cycle checkpoints.
b. Cancer cells have extra copies of chromosomes.
c. Cancer cells are oncogenes.
d. Cancer cells do not exhibit anchorage dependence.
110. The medium ground finch is one of Darwin’s finches that inhabit an island in the Galapagos. It
has a small beak that it uses to eat small, soft seeds. In 1977, a drought eliminated all but the
hard, large seeds. The birds who had larger beaks were able to find food and survive to
reproduce while the birds who had small beaks could not find food and died. Within a few
generations, the beak size of the medium ground finch increased by 10%. What type of
selection is this?
a. Disruptive
b. Stabilizing
c. Directional
d. Sympatric
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Final Review
Questions #111-112 refer to the following situation:
A plant found in Yosemite National Park is found to have two varieties: red flowers and blue
flowers. To determine the type of inheritance pattern found in this species, red flowers are
crossed with blue flowers.
111. What flowers would result if flower color has incomplete dominance?
a. Red flowers
b. Purple flowers
c. Blue flowers
d. Red and blue spotted flowers
112. What flowers would result if flower color has codominance?
a. Red flowers
b. Purple flowers
c. Blue flowers
d. Red and blue spotted flowers
113. An unknown lab sample is subjected to hydrolysis. The results include a combination of
hydrocarbon chains, glycerol molecules, and phosphate groups. What was in the original
sample?
a. Cholesterol
b. Phospholipids
c. Trans fats
d. Triglycerides
114. The cellular membrane is composed of a variety of proteins, each with different functions.
Which of the following is not a function of membrane proteins?
a. Energy storage
b. Transport
c. Signal transduction
d. Cell-cell recognition
115. After transcription, pre-mRNA undergoes processing that ensures, among other things, that it
will be safely transported to the cytoplasm and protected from hydrolytic enzymes. Which of
the following is not something that occurs during mRNA processing?
a. Poly-A tail
b. Recombination
c. 5’ cap
d. Splicing
116. Crossing over was first described in 1909 by Belgium professor Frans Alfons Janssens. It is a
phenomenon that has important implications for evolutionary change in populations. What is
the result of this phenomenon?
a. It creates chromosomes with genetic combinations unique from parental chromosomes.
b. It creates chiasma between nonhomologous chromosomes.
c. It allows for independent assortment to occur.
d. It creates stability in recombinant chromosomes.
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Final Review
Questions #117-118 refer to the following food web:
Source: http://alaska.usgs.gov/science/biology/seabirds_foragefish/marinehabitat/index.php
117. Which organism is a producer in this food web?
a. Cephalopod
b. Phytoplankton
c. Fox
d. Zooplankton
118. If the salmon population crashes as a result of overfishing, which of the following is not a
plausible result?
a. The fox population may increase.
b. The zooplankton population may increase.
c. Puffins may eat larger amounts of sand lances.
d. The cephalopod population may increase.
119. Mendel did most of his work during the mid-1800s. He grew and observed over 5,000 pea
plants and, from these studies, identified the law of independent assortment and the law of
segregation. Evidence for the law of segregation was later uncovered in the late 1800s by other
scientists. What process would provide evidence for the law of segregation?
a. Mitosis
b. Crossing over
c. Meiosis
d. Epistasis
120. Which of the following statements regarding cellular respiration rate is correct?
a. At 10oC, a gecko will have a higher rate of cellular respiration than a rat.
b. At 10oC, a gecko and a mouse will have equal rates of cellular respiration.
c. At 25oC, a mouse will have a higher rate of cellular respiration than a human.
d. At 25oC, a human will have a higher rate of cellular respiration than a mouse.
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Final Review
121. When Charles Darwin visited the Galapagos Islands, he observed the marine iguana. He later
decided that the marine iguana evolved from the terrestrial iguana found on the mainland of
South America. The marine iguana differs from the terrestrial iguana in several ways. The
marine iguana has a flat tail, long claws and sharp teeth while the terrestrial iguana has a round
tail, short claws and flat teeth. The marine iguana became a separate species after being
geographically isolated on the Galapagos Islands for several years. What type of speciation is
this?
a. Adaptive
b. Allopatric
c. Sympatric
d. Geographic
122. Structure and function is a recurring theme in biology. How does the structure of tRNA relate to
its function?
a. The anticodon pairs with mRNA while the amino acid attached to the other end bonds
with the growing polypeptide chain.
b. The three loops of the tRNA molecule are each able to pair to a different amino acid,
making translation efficient.
c. Allosteric control of tRNA can occur at any of the three anticodon locations, thereby
regulating translation.
d. The three loops of the tRNA molecule allow it to fit snugly into each of the three
openings in the rRNA molecule.
123. During translation, the growing peptide chain resides in the P site of rRNA with the next amino
acid in the chain waiting in the A site. Erythromycin, a common antibiotic, prevents the tRNA at
the A site from moving into the P site. How does this result in the death of the bacteria?
a. This terminates the replication of DNA.
b. This prevents the translation of proteins necessary for life.
c. This prevents the generation of RNA from DNA.
d. This creates a breakdown in metabolism, resulting in starvation of the bacteria.
124. Eukaryotes include all of the following organisms, except for one. Which organism is not
eukaryotic?
a. Euglena, a unicellular flagellate protist
b. Methanogens, bacteria that produce methane and live in the intestines of other
organisms
c. Mold, a multicellular organism that often grows on the surface of food
d. Duckweed, an aquatic plant that grows on the surface of ponds
125. You are studying two variations of banana slugs. They have similar coloration and live in the
same habitat. Although they are physically able to mate, there are no offspring that survive
embryologic development. What can you conclude about the banana slugs?
a. Prezygotic barriers have forced them to be separate species.
b. They are in Hardy-Weinberg equilibrium.
c. Microevolution is occurring in both variations.
d. They are two separate species.
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Final Review
126. The image below illustrates the process of cytokinesis in animals (1) and plants (2). Why are
there two different methods for the same process?
Source: http://commons.wikimedia.org/wiki/File:Cytokinesis.png
a. The cell wall of a plant cell prevents the contraction necessary for cleavage.
b. Animals evolved after plants and therefore developed a more efficient method of
cytokinesis.
c. Plants undergo fission, a slightly different version of mitosis.
d. The cell plate in plants pushes outwards, resisting the formation of a cleavage furrow.
QUANTITATIVE
1. A triglyceride is formed from a glycerol molecule, C3H8O3, and three identical fatty acids,
C7H14O2. How many hydrogen molecules are in the triglyceride?
2. Suppose an artificial “animal-type” cell was constructed out of dialysis tubing and contained
0.5M of a NaCl solution. If the cell is stored at 273 degrees Kelvin, what is the water potential of
the cell?
3. A microrespirometer was used to collect data on the respiration of unsprouted seeds. Calculate
the rate of cellular respiration, in cm/min, during last five minutes of the experiment.
Total Time (min)
Total distance fluid has moved
(cm)
0
0
0
5
3.6
3.6
10
7.4
3.8
15
9.6
2.2
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PSI AP Biology
Change in fluid position (cm)
Final Review
4. The influenza virus is spherical in shape with a genome of eight different RNA molecules. It can
have a diameter of up to 200 nm. The animal cells that it infects have a cell diameter of about 10
µm. How big is the average animal cell in nm?
5. The illustration below refers to the results from gel electrophoresis. In the unknown sample,
what is the size of the DNA fragment that was closest to the positive electrode while the gel was
being run?
6. The hemoglobin found in adult humans is composed of four polypeptide subunits. There are two
α subunits, composed of 141 amino acids each, and two β subunits, composed of 146 amino
acids each. How many nucleotides compose the codons necessary to translate this important
protein? Do not include the nucleotides used in start and stop codons.
7. The common earthworm has 36 chromosomes. How many different ways can these
chromosomes line up along the metaphase plate during meiosis I?
8. In pea plants, flowers axial flower position (A) is dominant to terminal flower position (a),
inflated pods (I) are dominant to constricted pods (i) and purple flowers (P) are dominant to
white flowers (p). In a cross of pea plants that are heterozygous for all traits, how many of the
480 offspring are also heterozygous for all traits?
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Final Review
9. In fruit flies, wild type flies have gray bodies and normal wings. Mutant flies have black bodies
and vestigial wings. A test cross between wild type and mutant flies, results in the following:
 1,930 wild type
 1,888 mutant
 412 gray, vestigial
 370 black, normal
What is the recombination frequency for the traits body color and wing type?
10. A population of wildflowers consists of two flower colors. Yellow flowers (Cy) are completely
dominant to white flowers (Cw). If the frequency of the Cy allele is 70%, what is the frequency of
the Cw allele?
11. The table below shows the biomass per square meter of a pond ecosystem. What is the
production efficiency of the primary consumers?
Trophic Level
Tertiary consumers
Secondary consumers
Primary consumers
Primary producers
Biomass (kg/m2)
1.2
10
41
512
12. Kruger National Park is home to a population of elephants that are protected within the park
boundaries. In 1992, the elephant population was 603. By 2007, the elephant population had
grown to 3,158. If the population continues to grow at this same rate, in what year will the
population reach the park’s theoretical carrying capacity of 7,000?
FREE RESPONSE
1. Phospholipids are important components to living systems. The diagrams below show two
phospholipid models and 2 models of phospholipid aggregations (groupings).
a. When in the presence of water, phospholipids can aggregate into a planar bilayer or a liposome
formation. Describe how the structure of phospholipids leads to both of these formations.
b. Explain the importance of phospholipids in the origin of life.
http://www2.chemistry.msu.ed
u/faculty/reusch/VirtTxtJml/Ima
ges3/bilyrstr.gif
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Final Review
2. Plants lose water from their above ground surfaces in the process of transpiration. Most of this
water is lost from stomata, microscopic openings in the leaves. Excess water loss can have a
negative effect on the growth, development, and reproduction of a plant. Severe water loss can be
fatal. Environmental factors have a major impact on the rate of plant transpiration.
a. Use the structure of the water molecule to explain why transpiration rate increases from 20⁰C
to 27⁰C.
b. Why does transpiration rate decrease after 27⁰C?
3. Himalayan rabbits carry the C gene, which is required for the development of pigments in the fur,
skin, and eyes. The C gene is inactive above 35°C, and it is maximally active from 15°C to 25°C. The
rabbits below have the same alleles for the C gene.
a. Describe the environments each of these rabbits were reared in. Provide the evidence that
supports your choice of environment.
b. Describe the type of ecological factor that regulates pigmentation in Himalayan rabbits.
4. Ricin is a protein found in the seeds of the castor plant (Ricinus communis). It is one of the most
potent poisons of eukaryotic cells. As shown in the diagram, the pathway for internalization of ricin
involves endocytosis by coated pits/vesicles (1) or by smooth pits/vesicles (2). Some ricin molecules
are returned to the cell surface via exocytosis (3). Other molecules fuse with lysosomes where the
ricin is destroyed (4). If ricin vesicles fuse with the Trans Golgi Network (5) there is a chance that
they may return to the cell surface via exocytosis (6). Ricin toxicity occurs when ricin penetrates the
Trans Golgi Network membrane and is liberated into the cytosol (7). Once inside the cytosol, ricin
catalyzes the depurination (removal of a nucleic base from the 28S ribosomal RNA) of the
ribosomes, halting protein synthesis. http://www.ansci.cornell.edu/plants/toxicagents/images/cell.gif
a. How does impairment of ribosomal RNA halt protein synthesis?
b. Describe the type of immune response described in the steps 2-4 of the above diagram.
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Final Review
5. The crested phenotype in rock pigeons is defined as
neck and occipital feathers growing towards the
top of the head and not down the neck of the
pigeon. Some strains of the rock pigeon (Columba
livia) with the crested (figure B) or non-crested
phenotype (figure A) are shown. Figure C shows
variants for a part of the protein EphB2 sequence, a
kinase that is involved in producing both
phenotypes during embryonic development, in
different organisms.
Genomic Diversity and Evolution of the Head Crest in the Rock Pigeon,
Science 1 March 2013: 339 (6123), 1063-1067.Published online 31 January 2013
A. Non-Crested Rock Pigeon(+)
B. Crested Phenotype in Rock Pigeon(cr)
a. Describe how a change in just one letter in Figure C can result in a change in phenotype.
b. There are 350 different breeds of Columba livia, with 80 of them displaying the crested
phenotype. Does it make more sense that the EphB2 mutation evolved separately several
different times or that it evolved once and was passed on to all of the crested breeds? What
theory supports your answer?
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Final Review
6.
ATP Synthase
The diagram shows the integral membrane protein complexes responsible for
electron transport and proton translocation in oxygenic photosynthesis. The
structures are from thermophilic cyanobacterial sources. Lumen (p) and
stromal (n) -side soluble electron transfer proteins are plastocyanin (PC) or
cytochrome c6, ferredoxin (Fd), and ferredoxin-NADP+ reductase (FNR).
Structure of the Cytochrome b6f Complex of Oxygenic Photosynthesis: Tuning the Cavity
Genji Kurisu, Huamin Zhang, Janet L. Smith, and William A. Cramer
Science 7 November 2003: 302 (5647), 1009-1014
Rotary model of how the E. coli F1F0
ATP synthase catalyzes the
synthesis of ATP. The protonmotive force drives rotation of a
ring composed of 12 c subunits (10
in the yeast). Protons enter the
assembly through a periplasmic
inlet channel and bind to the Asp61
carboxylate (open circle) of the c
subunits. The protonated binding
site (filled circle) then moves from
the a1b2 stator component to the
lipid phase of the membrane. After
12 steps the protons reach an
outlet channel on the F1-binding
(cytoplasmic) side of the
membrane.
Molecular Rotary Motors
Robert H. Fillingame
Science 26 November 1999: 286 (5445), 16871688.
a. Describe how an absence of water would affect ATP generation.
b. How would the ATP synthase be oriented in above model?
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Final Review
7. These images show an erythroid (red blood cell) enhancer that promotes the expression of BCL11A.
This protein represses fetal globin genes (HBG1, HBG2) in adults. The images show the effects of
normal, mutated and deleted erythroid enhancers on sickle cell disease.
GWAS to Therapy by Genome Edits?
Ross C. Hardison and Gerd A. Blobel
Science 11 October 2013: 342 (6155), 206-207
a. Sickle cell disease occurs as a result of a mutation in the hemoglobin gene. In terms of protein
structure, describe how a single mutation could lead to the disease.
b. Sickle cell disease offers some resistance to malaria. Individuals with the sickle cell formation
have fewer symptoms and are better able to survive the disease. Sickle cell disease is hereditary,
following a codominant inheritance pattern. For individuals who live in an area with high levels
of malaria, would it be more advantageous to have sickle cell disease or to be heterozygous for
the trait?
8. The illustration below shows proteolysis (enzymatic cutting of protein chains) at the metaphaseanaphase transition. A model is presented for how anaphase-promoting complex (APC) participates
in the breakdown of both mitotic cyclins and anaphase inhibitors. The model summarizes findings
from various organisms. Although some substrates have been identified in only a single organism,
the APC appears to have a universal role in mitotic proteolysis in eukaryotes. In this model, cyclin B
accumulates during interphase to activate CDC2 (dotted arrow), producing an active mitotic cyclindependent kinase (CDK) that triggers entry into prophase, resulting in the eventual formation of the
mitotic spindle and metaphase plate. The mitotic CDK also leads to the activation of the APC by an
undefined pathway. Once the APC is activated, it catalyzes the ubiquitination (adding ubiquitin, a
small 76 amino acid protein, to other proteins) of several substrates, including the anaphase
inhibitors PDS1 (budding yeast) and CUT2 (fission yeast). Proteins that have been tagged with
ubiquitin are slated to be broken down by the proteasome (an organelle that cuts proteins and aids
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Final Review
in protein recycling among other functions). The APC also mediates proteolysis of cyclin B, which is
universally required for exit from telophase (cytokinesis).
a. If a mutation results in the inability of a cell to produce cyclin B, what will be the fate of that cell,
in terms of the cell cycle?
b. How does ubiquitination affect anaphase inhibitors? How does this affect the cell cycle?
9. The human large intestine
harbors a complex community of
microbiota that affect many
aspects of our physiology and
health. Germ-free mice
inoculated with microbiota from
obese or lean human twins take
on the microbiota characteristics
of the donor. Those receiving the
obese microbiota had an increase
in adiposity (fat cell size), whereas
those receiving the lean
microbiota remained lean.
Gut Microbiota from Twins Discordant for Obesity
Modulate Metabolism in Mice
Vanessa K. Ridaura, Jeremiah J. Faith et. al.
Science 6 September 2013: 341 (6150), 1241214
a. Describe a control group that
could have been used for this
experiment?
b. Using the graph labeled E,
which mice had greater
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Final Review
changes in body mass- and what do the larger errors on the top curve (obese twin) than the
bottom curve mean?
10. Waardenburg syndrome (WS) is an inherited birth defect in which a baby
will suffer from hearing loss and problems with the pigment in the skin and
hair. It is an autosomal dominant disease.
The hearing loss in WS is present from birth (congenital). The hearing loss is
typically not progressive, can be in one or both ears, and of the
sensorineural (nerve) type. The hearing loss can vary from moderate to
severe to profound. The following are additional signs and symptoms that
may be seen in WS, noting that the child may not have all of these
characteristics. The pigment (color) of the eyes and skin are affected
causing two differently colored eyes or white areas of the skin (albinism).
The bridge of the nose may be wide and the eyes may appear to be spaced far apart. The hairline
may be low on the forehead. Hair may become gray early in life, often before 30 years of age. There
may be patches of white hair such as a “white forelock”, which is a white lock of hair growing above
the forehead. Type I WS includes the above signs and symptoms. Type II WS also has the above signs
but has hearing loss and differently colored eyes more commonly than in type I. On the other hand,
the white forelock, early graying of the hair, and albinism occur less commonly than in type I. Type
III has the above signs, plus malformations of the upper limbs. Type IV has the above signs, as well
as Hirschsprung disease, which is a digestive disorder where certain nerve cells in the large intestine
are missing that causes problems passing stool.
www.i-am-pregnant.com
a. A woman with WS marries a man whose family has no history of the disease. They have a
daughter with WS and two normal phenotype sons. One son marries a woman whose family
has no history of the disease. They have two daughters. Create a pedigree for this family.
b. What is the probability that the granddaughters will have the disease? Suppose that the
daughter of the original couple marries a man with no family history of WS. What is the
probability that they will have a child affected with the disease?
11. Double-gene knockout (removal of DNA or its functionality) mice (V1a–/–V1b–/–) and wild-type
mice (WT) are examined for their behavioral rhythms under experimental jet-lag conditions. The
mice were housed in a light-controlled isolator with food and drink available, and their body
movement activity was recorded by infrared (heat) sensors.
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The small, dark bands in the above figures are evidence of activity or movement by the mice. When they
were maintained in a 12-hour light/12-hour dark (LD) cycle, both WT and V1a–/–V1b–/– mice exhibited
high locomotor activity during the dark phase (12-0 hours). After 2 weeks of recording behaviors (-5 to 0
days), LD cycles were advanced by 8 hours. In WT mice, this advance evoked a gradual shift of locomotor
activity rhythms, which took 8 to 10 days for complete adjustment to the new LD schedule. This slow
resetting of locomotor activity rhythm was expected, as it is the typical sign that mice are experiencing
jet lag. Every subsequent day after the LD cycle advance, the WT mice will start their activity slightly
earlier, to finally align, after 8 to 10 days, to the beginning of the night. In contrast, V1a–/–V1b–/– mice
showed almost immediate readjustment within only 2 to 4 days of transition.
a. Suppose that an equal number of each type of mice is released into an ecosystem that
experiences random fluctuations of day and night. How would natural selection alter the alleles
in the population of mice?
b. In this new ecosystem, the fluctuations of day and night cause an unstable environment, in
which the mice have relatively short lives. Describe the type of survivorship strategy that would
be used by the mice in this environment.
12. These graphs show the results of laboratory feeding experiments of the snail T. funebralis. The
experiment tested the feeding preference of
the snails for the vegetative blades (V) and
sporophylls (S) of the brown seaweed A.
marginata.
Graph A
Ten large (wet weight, about 10 g) T.
funebralis (snails) were offered equal weights
(about 10 g each) of A. marginata sporophylls
and vegetative
blades in 36-liter aquaria. After 48 hours, the
algae were weighed again. Control aquaria
contained algae but no snails.
Graph B
For total phenolic content and tanning ability
measurements, fresh algal tissue was ground
in 50 % methanol in a tissue macerator and
extracted in the dark for 24 hours.
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Final Review
Graph C
Organic nitrogen content refers to proteins and
amino acids.
Graph D
The toughness of the blades and sporophylls
refers to how easy it can be cut or punctured.
Algal Chemical Defense Against Herbivores: Allocation of Phenolic Compounds in the Kelp Alaria marginata
PETER D. STEINBERG Science 27 January 1984: 405-407
a. Based on the data, which part of brown seaweed is preferred by the snails? Describe your
rationale.
b. During the experiment, why was the algae weighed a second time, 48 hours after placement in
the aquaria?
13. The Miller-Urey experiment showed amino acids can be produced from gases in the Earth’s early
atmosphere and environmental conditions (heat and lightning). These amino acids could then go on
to be used as reactants in the synthesis pathway for RNA and DNA. This evolution of genetic
material allowed the progression of life on Earth.
a. It is highly likely that amino acids evolved before either RNA or DNA. Which molecule most likely
followed amino acid evolution? Explain your reasoning.
b. The classic Miller-Urey experiment tested the hypothesis that organic molecules such as amino
acids could be created from lightning in the early atmosphere. Describe two other possible
sources of organic molecules.
c. In the evolution of prokaryotic cells into eukaryotic cells, several organelles are thought to have
been taken in by other cells along with their DNA. Name one of these organelles and explain this
theory.
14. During microscopic examination of human tissue samples, a student observed the following.
a. The tissue in this sample is ideally suited for the
exchange of materials with the environment.
Propose a location in the human body where
this tissue may be found and describe its
function.
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b. How does the structure of the above tissue relate to its function of material exchange?
c. Suppose that three blocks are composed of agar, phenolphthalein and base solution. They are
cut into differing sizes (1 cm3, 2 cm3 and 3 cm3) and placed in an acid solution for a set amount
of time. After the blocks are retrieved from the acid solution, they are cut in half to measure the
distance of diffusion. Describe the results expected. How does size relate to diffusion rate?
15. The diagram shows a phylogenetic tree made from the amino acid sequences of the complete
haemocyanin molecules of Sepia officinalis (SoH1-SoH3, cuttlefish), Enteroctopus dofleini (OdHA,
OdHG, Giant Pacific octopus), Nautilus pompilius (NpH, nautilus) and Haliotis tuberculata (HtH1,
HtH2, green ormer (sea snail)). The split of gastropoda (snails and slugs) and cephalopoda (mollusks)
about 550 ± 50 millions of years ago (mya) was used as calibration point to estimate the DNA
sequence mutation rate. Included are the exon-intron (rectangles-line) structures of the
haemocyanin of N. pompilius, E. dofleini and S. officinalis. Grey bars indicate the 95% HPD (highest
posterior density) or the confidence interval of the estimated age. The number under the root of
each branch is the estimated age of gene duplication in mya. Haemocynanin is an oxygen-binding,
cylindrical protein complex made from at least 10 subunits.
a. If you would look at the sequence alignment of the sequences used to make this
phylogenetic tree, which 2 sequences would be the most similar? Describe the evidence
to support your choice.
b. The exon-intron structure of the Haliotis tuberculata haemocynanins would be expected
to be more similar to which species? Describe the evidence to support your choice.
c. This phylogenetic tree is based on what type of homologies? Name and describe one
other type of homology that could be used to build a phylogenetic tree.
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16. Spartina anglica is a salt marsh grass found along the coasts of the British Isles. S. anglica is a very
tolerant plant that can withstand periodic inundation and water-saturated soils. Because of these
adaptations, it is used extensively for stabilizing mudflats. In general, S. anglica is the most seaward
grass found in a salt marsh.
Molles, Manuel C. "Competition." Ecology: Concepts and Applications. Boston: WCB/McGraw-Hill, 1999. 237. Print.
The distribution of S. anglica in a salt marsh depends on several factors. Tidal fluctuations determine the
upper and lower limits of distribution, with the grass being found between the mean high water spring
tide and the mean high water neap tide. A second factor that determines distribution is the fetch of the
estuary. The larger the fetch (distance over which wind can blow) the higher S. anglica must live in order
to avoid wave disturbances. In the upper intertidal zone, S. anglica are replaced by other marsh grasses
which are better able to compete for resources in those areas of the marsh.
a. The distribution of S. anglica at the upper and lower levels of the intertidal zone are determined
by different factors. For each location, describe the environmental factors affecting S. anglica as
biotic or abiotic and density-dependent or density-independent.
b. Describe the interaction occurring in the upper intertidal zone between S. anglica and other
marsh grasses.
c. To further investigate the interaction between S. anglica and other marsh grasses, an exclusion
experiment is run. Describe how this would work, including possible results.
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ANSWER KEY
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
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27.
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32.
33.
34.
35.
36.
37.
38.
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40.
41.
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43.
44.
45.
46.
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D
A
C
D
A
C
B
A
C
C
C
A
D
A
C
D
A
D
C
D
A
B
C
A
C
B
C
B
D
C
A
C
D
D
B
D
B
C
B
C
A
C
B
A
D
B
47.
48.
49.
50.
51.
52.
53.
54.
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56.
57.
58.
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60.
61.
62.
63.
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80.
81.
82.
83.
84.
85.
86.
87.
88.
89.
90.
91.
92.
B
C
B
C
D
B
A
A
D
B
D
C
B
C
A
B
C
D
B
D
D
B
D
D
C
C
B
B
A
B
C
A
B
D
A
A
D
B
A
B
D
A
A
B
A
C
PSI AP Biology
93.
94.
95.
96.
97.
98.
99.
100.
101.
102.
103.
104.
105.
106.
107.
108.
109.
110.
111.
112.
113.
114.
115.
116.
117.
118.
119.
120.
121.
122.
123.
124.
125.
126.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
D
C
D
D
C
B
A
A
D
B
A
D
B
B
B
D
A
C
B
D
B
A
B
A
A
A
C
C
B
A
B
B
D
A
44
22.7 bars
0.44 cm/min
10,000 nm
1500 bp
1722
6.9 x 1010
60
17%
30%
8%
2014
Final Review
1. Phospholipids
a. Water molecules are polar. The phosphate head of a phospholipid is polar while the tail is
nonpolar. The head is attracted to water while the tail is repelled by water. This causes the
heads to aggregate beside water molecules while the tails form a middle layer.
b. Phospholipids made it possible for the existence of micro-environments that were isolated
form an outer environment. In these micro-environments, evolution was able to occur.
2. Transpiration
a. Water molecules are held strongly together via hydrogen bonds. In order for water to
evaporate, energy must first go into breaking the hydrogen bonds. As temperature
increases, the amount of hydrogen bonds that are broken also increases, allowing
transpiration rate to increase.
b. At temperatures above 27⁰C, stomata begin to close to conserve water, leading to a lower
transpiration rate.
3. Rabbits
a. The rabbit with the black tips was raise in an environment where the temperature was
between 15°C to 25°C, while the totally white was reared in an environment above 35°C.
The C gene is inactivate above 35°C and therefore would be able to affect the color of the
fur in the defined regions of the rabbit.
b. Temperature is the factor that regulates pigmentation in Himalayan rabbits. It is an abiotic
factor.
4. Ricin
a. Ribosomal RNA is the location where nucleotides are joined to a growing polypeptide chain.
By impairing the function of rRNA, protein synthesis would no longer occur.
b. Phagocytosis is a part of innate immunity. Phagocytosis occurs when pathogens are taken
into the cell via endocytosis. Vesicles with the pathogen are fused with lysosomes where the
pathogen is destroyed.
5. Pigeons
a. The sequence of letters represents the amino acid sequence of protein variants. A change in
one amino acid results in a change in the folding of that protein, resulting in different
function and, therefore, phenotype.
b. The theory of maximum parsimony supports the idea that the mutation occurred once and
was passed on to all of the crested breeds that followed.
6. Photosynthesis
a. Water provides the stream of electrons that are given to photosystem II to create the
proton gradient to generate ATP and the reducing powering in NADPH. Without water, this
ATP generation will fall.
b. The ATP synthase would be situated in the membrane with rotors facing up or into the
lumen of thylakoids and the ATP-binding site facing out or into the stroma of the
chloroplast.
7. Hemoglobin
a. The single mutation causes a change in a codon for an amino acid in the synthesis of
proteins related to red blood cells. Amino acid sequence is the primary structure of proteins.
A change at this level of structure affects all other levels, causing conformational changes in
the final protein.
b. It would be best to be heterozygous. This would allow some resistance to malaria while also
giving the individual a mixture of healthy and sickle cell hemoglobin instead of all sickle cell
hemoglobin.
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8. Cyclins
a. Cyclin B is necessary to trigger entry into prophase. Without an accumulation of cyclin B, the
cell will remain in the G2 phase and the cell will not divide.
b. Ubiquitination causes the anaphase inhibitors to be broken down. This means that anaphase
is not inhibited and the cell will continue through the cell cycle and divide successfully.
9. Twin Mice
a. A control group would have been mice that received neither microbiota treatment.
b. The obese twin pair inoculations had greater changes in fat mass than the lean inoculations.
The larger error bars on the curve show that some mice show more fat mass gain than other
mice that gain fat mass. The larger error bars signify that possibly there other factor that
contribute to body mass index, such as bacterial strain fluctuations in the composition of the
microbial population.
10. Heredity
a.
b. 0%; 50%
11. Mice
a. The mutated mice would have an advantage because they are able to adapt faster to
changing conditions. These mice would survive to reproduce more so than the wild type
mice. In the following generations, a majority of alleles will be provided by the mutated
mice.
b. The mice would use r-selection (or Type III survivorship) survivorship strategies. They would
have high fecundity, early maturation and low parental care.
12. Snails
a. The snails prefer the vegetative blade of the brown kelp. The vegetative blade has a much
lower level of phenols.
b. In order to determine the amount of algae eaten by the snails, the weight of the algae
would need to be recorded before placement and after placement in the aquaria. The after
value would be subtracted from the before value to calculate the amount eaten.
13. Miller Urey
a. RNA probably evolved before DNA. Although both molecules can store genetic information,
RNA can catalyze reactions and can reproduce itself and create DNA.
b. Organic molecules could also be created via volcanic eruptions on early Earth. Other
theories postulate that meteorites brought important molecules to Earth.
c. Chloroplasts and mitochondria are thought to have been free-living bacteria that were taken
in by other cells. Known as the endosymbiotic theory, the DNA of the bacteria were
incorporated into the larger cell’s genome and eventually became a permanent part of that
cell.
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14. Human Tissue
a. Answers may vary. Example: the lining of the small intestine where nutrients from food
diffuse into blood vessels to be carried throughout the body.
b. The folds of the tissue increase the surface area to volume ratio of the tissue. This allows a
maximum amount of materials to be exchanged with the environment in a given amount of
time.
c. The smallest block (1cm3) will show the highest diffusion rate, followed by the 2cm3 block
and then the 3cm3 block. The 1cm3 block has the highest surface area to volume ratio while
the 3cm3 block has the lowest ratio. The higher the surface area to volume ratio, the faster
the rate of diffusion.
15. Phylogenetics
a. Enteroctopus dofleini (OdHA, OdHG, Giant Pacific octopus) sequences would be most
similar, because the time from their duplication is the smallest; phylogenetic trees are
calibrate on amount of mutations between all sequences involved in the study.
b. The exon-intron structure of the Haliotis tuberculata haemocynanins would be most similar
to Nautilus pompilius (NpH, nautilus). The nautilus heamocyanin gene has the g1g2 and s1s2
exon duplications like the sequence leading to the mollusks but does not have the exons b,
e, f duplicated.
c. This tree is based on molecular homology. Structural homology could also be used. This
compares the anatomical structure of the animal, both as an adult and as an embryo.
16. Marsh
a. Upper intertidal zone: biotic, density-dependent; Lower intertidal zone: abiotic, densityindependent
b. Competitive exclusion is occurring between S. anglica and other marsh grasses. Other
grasses are better able to compete for resources, thereby excluding S. anglica from growing
there.
c. Answers may vary slightly. For an exclusion experiment, upper intertidal plots would be built
and studied. 1) a plot with only S. anglica. 2) a plot with both S. anglica and another
common grass. If S. anglica grows in the exclusion plot but not in the combined plot, then
this shows that the other grass is a better competitor for resources. If S. anglica does not
grow well in either plot, then the limiting factor is environmental and not due to the other
grass.
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